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NCERT Solutions for Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data (PDF) – 2027

NCERT Solutions for Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.

Chapter 10 asks what happens to an average when groups of data, mixtures or marks are combined. Averaging the averages works only when the groups are the same size; otherwise each part must count according to its size or importance, which gives the weighted mean (Brahmagupta, c. 628 CE). The second half turns to pictures of combined data: stacked bar charts, which show totals and their parts, and 100% stacked bar charts, which show shares. You will learn to compute weighted means and to judge what each chart can and cannot tell you.

NCERT Solutions for Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data PDF

All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.

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NCERT Solutions for Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data – Questions and Answers

All 60 questions of the chapter (textbook pages Part II, 8 – 36) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.

Think and Reflect — Page 8Section 10.1.1 Average of Averages (Example 1)

Think and Reflect
Q1In a badminton academy there are 11 trainees — 8 seniors and 3 juniors. Their heights (in cm) are:
Heights (in cm)Average Height (in cm)
Seniors165, 169, 164, 167, 170, 159, 164, 166165.5
Juniors146, 149, 153149.33
Method 1 found the average height of the whole group as (165.5 + 149.33) ÷ 2 = 157.415 cm. Method 2 added all eleven heights and divided by 11 to get 1772 ÷ 11 = 161.09 cm, which is correct.
Why did Method 1 not work?
Answer

Method 1 gives the two group averages equal importance, as if there were as many juniors as seniors. But there are 8 seniors and only 3 juniors. Each senior’s height should count once, so the seniors’ average must count 8 times and the juniors’ average only 3 times.

Correct average = (165.5 × 8 + 149.33 × 3) ÷ (8 + 3)
= (1324 + 448) ÷ 11 = 1772 ÷ 11 ≈ 161.09 cm

Method 1 pulls the answer down towards the juniors’ average, so it is too small (157.4 cm).

Exercise Set 10.1 — Page 13Sections 10.1.1 Average of Averages and 10.1.2 Mixtures

Q1The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students?
Answer
Total score of Section A = 72 × 30 = 2160
Total score of Section B = 76 × 25 = 1900
Combined average = (2160 + 1900) ÷ (30 + 25)
= 4060 ÷ 55 ≈ 73.82

The combined average is about 73.82. It is closer to 72 than to 76 because Section A has more students.

Q2A farmer mixes three equal quantities of fertilisers. The first one contains 1⁄10 nitrogen, the second contains 9⁄50 nitrogen, and the third contains 3⁄60 nitrogen. What is the fraction of nitrogen in the mixture?
Answer

The quantities are equal, so the fraction in the mixture is the simple average of the three fractions.

1⁄10 = 30⁄300, 9⁄50 = 54⁄300, 3⁄60 = 15⁄300
Sum = 99⁄300 = 33⁄100
Fraction of nitrogen = 33⁄100 ÷ 3 = 11⁄100 (that is, 11%)
Q3(Śrīdharācārya, Pāṭīgaṇita, c. 750 CE) In ancient India, Varṇa was the measure of gold purity. A purity of 16 varṇa meant pure gold; in general, a purity of k varṇa meant that the gold-alloy was k/16 gold and the rest impurities. (Now the term used is karat; 16 Varṇa = 24 karat.)
Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varṇa, 5 units at 10 varṇa, and 17 units at 11 varṇa. Find the purity in varṇa of the combined gold.
Answer

Purity of the mixture is the weighted mean of the purities, with the amounts as weights.

Purity = (9 × 12 + 5 × 10 + 17 × 11) ÷ (9 + 5 + 17)
= (108 + 50 + 187) ÷ 31
= 345 ÷ 31 = 114⁄31 ≈ 11.13 varṇa
Q4The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.
Answer

May and July have 31 days and June has 30 days, so each month’s average is weighted by its number of days.

Combined average = (3.5 × 31 + 10 × 30 + 8.7 × 31) ÷ (31 + 30 + 31)
= (108.5 + 300 + 269.7) ÷ 92 = 678.2 ÷ 92
≈ 7.37 mm per day
Q5Calculate the concentration of spice mix in these two scenarios.
(i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 300 mL one with 15% spice mix are combined.
(ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 100 mL one with 15% spice mix are mixed.
Answer
(i) Spice = 100 × 0.05 + 200 × 0.10 + 300 × 0.15 = 5 + 20 + 45 = 70 mL
Concentration = 70 ÷ 600 ≈ 11.67%

(ii) Spice = 300 × 0.05 + 200 × 0.10 + 100 × 0.15 = 15 + 20 + 15 = 50 mL
Concentration = 50 ÷ 600 ≈ 8.33%

The same three strengths give different results because the amounts are swapped: in (i) most of the drink is 15%, in (ii) most of it is 5%.

Exercise Set 10.2 — Pages 14–15Section 10.1.3 Custom Weights

Q1Savitri’s marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in the project, and 80 out of 100 in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3 : 4 : 5. Which of the following expression(s) gives her annual score (as a percentage) in Kashmiri?
(i) (35 × 3 + 44 × 4 + 80 × 5) ÷ (3 + 4 + 5)

(ii) [35⁄100 × 3 + 44⁄100 × 4 + 80⁄100 × 5] ÷ (3 + 4 + 5)

(iii) {[35⁄50 × 3 + 44⁄60 × 4 + 80⁄100 × 5] ÷ (3 + 4 + 5)} × 100

(iv) [(35⁄50 × 100) × 3 + (44⁄60 × 100) × 4 + (80⁄100 × 100) × 5] ÷ (3 + 4 + 5)
Answer

(iii) and (iv) are correct.

First change each mark to a percentage of its own maximum: 35/50 = 70%, 44/60 ≈ 73.33%, 80/100 = 80%. Then take the weighted mean with weights 3, 4, 5.

Annual score = (70 × 3 + 73.33 × 4 + 80 × 5) ÷ 12
= (210 + 293.33 + 400) ÷ 12 ≈ 75.28%

(i) and (ii) are wrong: they treat 35 and 44 as if they were out of 100, but they are out of 50 and 60. (ii) also gives a fraction, not a percentage.

Exercise Set 10.3 — Pages 15–18Section 10.1 Combining Things: weighted mean in different situations

Q1A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?
Answer
Profit on books = 30% of ₹8000 = ₹2400
Profit on covers = 50% of ₹1000 = ₹500
Total profit = ₹2900 on total sales of ₹9000
Profit percentage = 2900 ÷ 9000 × 100 ≈ 32.22%

It is close to 30% because most of the sales (₹8000 of ₹9000) came from books.

Q2A white stork’s migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.
Answer

Guess: a little more than 44.5 km, because one day out of 21 is above the average.

Distance in 20 days = 44.5 × 20 = 890 km
Distance in 21 days = 890 + 55 = 945 km
New average = 945 ÷ 21 = 45 km per day
Q3A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?
(i) Salt: 5%, Sugar: 8%   (ii) Salt: 13%, Sugar: 3%
(iii) Salt: 6%, Sugar: 6%   (iv) Salt: 5.55%, Sugar: 8.88%
(v) Salt: 3.33%, Sugar: 2.67%   (vi) Salt: 4.1%, Sugar: 7.08%
Answer

The correct option is (v).

Salt = 5% of 600 = 30 mL; sugar = 8% of 300 = 24 mL
Total mixture = 600 + 300 = 900 mL
Salt = 30 ÷ 900 × 100 ≈ 3.33%
Sugar = 24 ÷ 900 × 100 ≈ 2.67%

Each substance is now spread through a larger volume, so both concentrations fall.

Q4At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to (3⁄4)th of the original concentration?
Answer
Spice in the pani = 8% of 10 L = 0.8 L
Required concentration = 3⁄4 × 8% = 6%
Let w litres of water be added (water has 0% spice):
0.8 ÷ (10 + w) = 0.06
10 + w = 0.8 ÷ 0.06 = 13.33…
w = 3.33… = 10⁄3 litres ≈ 3.33 L
Q5A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4 : 5 : 6. Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively.
(i) Find out whose total is more without calculating.
(ii) What are their final marks?
Answer

(i) Keerthi’s. Both have 65 in flexibility. Keerthi has 5 less in strength (weight 4) but 5 more in agility (weight 6). Agility has the larger weight, so her gain outweighs her loss.

(ii) Rashi = (60 × 4 + 65 × 5 + 70 × 6) ÷ 15 = (240 + 325 + 420) ÷ 15
= 985 ÷ 15 ≈ 65.67
Keerthi = (55 × 4 + 65 × 5 + 75 × 6) ÷ 15 = (220 + 325 + 450) ÷ 15
= 995 ÷ 15 ≈ 66.33
Q6A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown in the table below (number of customers giving each rating).
5★4★3★2★1★
Food53200
Ambience04510
Service12241
What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
Answer
Food = (5×5 + 4×3 + 3×2) ÷ 10 = 43 ÷ 10 = 4.3
Ambience = (4×4 + 3×5 + 2×1) ÷ 10 = 33 ÷ 10 = 3.3
Service = (5×1 + 4×2 + 3×2 + 2×4 + 1×1) ÷ 10 = 28 ÷ 10 = 2.8
Combined = (4.3 × 6 + 3.3 × 5 + 2.8 × 4) ÷ (6 + 5 + 4)
= (25.8 + 16.5 + 11.2) ÷ 15 = 53.5 ÷ 15 ≈ 3.57

The restaurant’s average rating is about 3.57 out of 5.

Q7The following table shows the weight data of langurs in an animal facility. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number?
Male langursFemale langursAll langurs
Average weight16.5 kg13.8 kg14.925 kg
Number of langurs60
(i) Which of the following expression(s) describes the given scenario?
(a) (16.5x + 13.8y) ÷ (x + y) = 14.925
(b) (16.5x + 13.8y) ÷ 60 = – 14.925
(c) (16.5x + 13.8y) ÷ 2 = – 14.925
(d) (16.5x + 13.8y) ÷ (16.5 + 13.8) = – 14.925
(ii) Find out how many male langurs are present.
(iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this?
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released from the facility. What is the average weight of the male langurs after this?
(v) Now, suppose one of the male langurs lost 1 kg of weight. What is the average weight of all the male langurs after this?
Answer

More females. The overall average 14.925 kg is closer to the female average (13.8) than to the male average (16.5), so the females carry more weight in the mean.

(i) (a) is correct (x males, y females). Since x + y = 60, (b) is also correct if read as “= 14.925”. (c) and (d) are wrong.

(ii) 16.5x + 13.8(60 − x) = 14.925 × 60 = 895.5
828 + 2.7x = 895.5 ⇒ x = 25 males (35 females)
(iii) (35 × 13.8 + 15.2) ÷ 36 = 498.2 ÷ 36 ≈ 13.84 kg
(iv) (25 × 16.5 − 16.9 − 16.1) ÷ 23 = 379.5 ÷ 23 = 16.5 kg
(v) (379.5 − 1) ÷ 23 = 378.5 ÷ 23 ≈ 16.46 kg

So there are 25 male langurs.

Q8Dorjee has collected 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions. A calculator can be used if necessary.
Water SourceSalinity
Dead Sea≈ 34%
Other seas and oceans≈ 3.5%
Ground water≈ 0.01%
Purified drinking water≈ 0.001%
(i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with 2 litres of purified drinking water?
(ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? Why/Why not? What quantity of purified drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater?
(iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not? What quantity of groundwater should he mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of purified drinking water?
Answer
(i) (1 × 34 + 2 × 0.001) ÷ 3 = 34.002 ÷ 3 ≈ 11.33%

(ii) Yes, because 0.01% lies between 0.001% and 34%. Let w litres be added:

(34 + 0.001w) ÷ (1 + w) = 0.01
34 + 0.001w = 0.01 + 0.01w
33.99 = 0.009w ⇒ w ≈ 3777 litres

(iii) No. A mixture’s salinity always lies between the salinities of the parts, here between 0.01% and 34%. It can never fall to 0.001%, which is below both. So no quantity of groundwater works.

In-text Questions — Pages 18–20Section 10.2.1 Stacking Columns (cluster-column charts, Fig. 10.1)

In-text
Q1The table shows the average weekly expenditure, in rupees (₹), of three families across categories.
HousingFoodEducationTransportationHealthcareRecreation
Family A19401700128012001770535
Family B175015461500128012100
Family C9501700154014001300150
Table 10.2, page 18
Choice 1: Comparing expenditure of families across categoriesFamily AFamily BFamily C0500100015002000HousingFoodEducationTransportationHealthcareRecreationChoice 2: Comparing category-wise expenditure of each familyHousingFoodEducationTransportationHealthcareRecreation0500100015002000Family AFamily BFamily C
Fig. 10.1, page 19 — Two cluster-column graphs showing different ways of visualising the data in Table 10.2 (expenditure in ₹)
Answer each question by looking at Choice 1 first, and then Choice 2.
1. Which family spent the least on housing?
Answer

Family C (₹950). In Choice 1 the three housing bars stand side by side and Family C’s bar is clearly the shortest (Family A ₹1940, Family B ₹1750).

Q22. Which family spent the most on healthcare? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer

Family A (₹1770, against ₹1210 for Family B and ₹1300 for Family C). Choice 1 shows this at a glance: Family A’s healthcare bar is the tallest in the healthcare cluster.

Q33. Approximately how much did Family A spend on education? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer

About ₹1300 (the table gives ₹1280). On either chart the education bar for Family A ends a little above the 1000 line, roughly halfway to 1500.

Q44. Which category did Family B spend most on? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer

Housing (₹1750). In Choice 2 all six of Family B’s bars are together, and the housing bar is the tallest; food (₹1546) and education (₹1500) come next.

Q55. Did Family A spend more on food or on healthcare? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer

On healthcare, but only slightly: ₹1770 on healthcare against ₹1700 on food. Choice 2 makes this easier, since both bars are in Family A’s cluster.

Q1–5Try to answer each of Questions 1–5 by looking at Choice 1 first, and then Choice 2. Which was easier?
Answer

Choice 1 was easier for Questions 1 and 2, because they compare families within one category, and Choice 1 puts those bars side by side.

Choice 2 was easier for Questions 4 and 5, because they compare categories within one family, and Choice 2 keeps each family’s bars together.

Question 3 asks for one value, so both charts work equally well.

Q66. Which family spent the least amount overall? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer
Family A: 1940 + 1700 + 1280 + 1200 + 1770 + 535 = ₹8425
Family B: 1750 + 1546 + 1500 + 1280 + 1210 + 0 = ₹7286
Family C: 950 + 1700 + 1540 + 1400 + 1300 + 150 = ₹7040

Family C spent the least overall (₹7040).

Q77. Approximately how much did Family C spend overall? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer

About ₹7000 (exactly ₹950 + ₹1700 + ₹1540 + ₹1400 + ₹1300 + ₹150 = ₹7040).

Q88. Approximately what percentage of Family C’s expenses were on housing? (Use Table 10.2 and Fig. 10.1 (shown in Q1 of this section).)
Answer
Housing ÷ total = 950 ÷ 7040 × 100 ≈ 13.5%

About 13.5% (roughly one-seventh) of Family C’s expenses were on housing.

In-text Questions — Page 20Section 10.2.1 Stacking Columns (stacked bar chart, Fig. 10.2)

In-text
Q6–8Was it as straightforward to answer Questions 6–8 (least total spending, Family C’s total, and Family C’s percentage on housing) as the earlier questions? Is there a visualisation that makes it simpler to answer such questions? Look at the following visualisation of the same data. Try to understand how the chart is organised and answer the three questions above.
Expenditure of three families in a monthHousingFoodEducationTransportationHealthcareRecreation0100020003000400050006000700080009000Family AFamily BFamily C
Fig. 10.2, page 20 — A stacked bar chart of the data in Table 10.2 (₹)
Answer

No, the cluster charts made us add six values per family. The stacked bar chart (Fig. 10.2) makes it simpler: each family’s bars are joined end to end, so the length of the whole bar is the family’s total.

6. Family C’s bar is the shortest, so Family C spent the least.
7. Family C’s bar ends at about ₹7000.
8. The housing (red) piece is about 950 of the 7000-long bar, roughly one-seventh, about 13–14%.

Q1–5Now, use the stacked bar chart (Fig. 10.2) to answer Questions 1–5 shown earlier. Is it possible? Is it easy? If not, what made it harder than before?
Answer

It is possible but harder.

Only the first piece (housing) starts at 0. Every other piece starts where the previous one ends, so to read, say, Family A’s education you must subtract the start from the end (about 4920 − 3640 = 1280). Also, the same category starts at different points in different bars, so comparing, for example, healthcare across families means comparing pieces that are not lined up.

Q1 (housing) is still easy, since housing starts at 0 in every bar. The others take more effort. Stacked bars make totals easy but individual parts harder.

Exercise Set 10.4 — Pages 21–22Section 10.2.1 Stacking Columns

Q1The following stacked column chart shows the number of animal species in the IUCN Red List, by class, over time.
The Number of Endangered Species is RisingNumber of animal species of the IUCN Red List, by classMammalsReptilesBirdsInsectsAmphibiansMolluscs (e.g., snails)FishOthers*03,0006,0009,00012,00015,0007,85120079,618201011,212201312,630201614,2342019
Ex. 10.4 Q1, page 21 (redrawn; segment heights are approximate, read from the printed chart; the totals are as printed). * Other invertebrate (spineless) animals, such as crustaceans, corals and arachnids (spiders, scorpions). Source: IUCN Red List
The International Union for Conservation of Nature (IUCN) Red List is a large, regularly-updated database that tells us the extinction risk of species across the world. It classifies species into groups such as Least Concern, Vulnerable, Endangered, Critically Endangered, Extinct in the Wild and Extinct, based on scientific data about their numbers, trends and habitats.
(i) What does the number 14,234 on the top of the column 2019 represent?
(ii) Approximately how many reptile species were in the list in 2016?
(iii) Which class(es) of species have seen a relatively small increase in count between 2007 and 2019?
Answer

(i) 14,234 is the total number of animal species (all classes together) on the endangered list in 2019, that is, the full height of the 2019 column.

(ii) About 1,000 reptile species. The dark red reptile piece in 2016 is about one-third of the gap between 0 and 3,000.

(iii) Mammals, birds and amphibians. Their pieces are only a little taller in 2019 than in 2007 (a rise of roughly a tenth to a quarter), while reptiles, insects, molluscs, fish and others roughly doubled or more.

Q2The wickets taken by a bowler in International Cricket matches till 2025 are shown in the table.
TestODIT20
Home6210032
Overseas1724971
Complete the given stacked bar charts (the bar lengths can be approximate),
(i) comparing the total wickets taken at home vs. overseas:
(ii) the total wickets taken in each format:
Wickets taken at home vs. overseasTestODIT20050100150200250300HomeOverseas
(i) To complete: one stacked bar each for Home and Overseas
Wickets taken in different formatsHomeOverseas050100150200250300TestODIT20
(ii) To complete: one stacked bar for each format
Answer
(i) Home: 62 + 100 + 32 = 194; Overseas: 172 + 49 + 71 = 292
(ii) Test: 62 + 172 = 234; ODI: 100 + 49 = 149; T20: 32 + 71 = 103
Wickets taken at home vs. overseasTestODIT20050100150200250300Home6210032Overseas1724971
(i) Completed: Home total 194, Overseas total 292
Wickets taken in different formatsHomeOverseas050100150200250300Test62172ODI10049T203271
(ii) Completed: Test 234, ODI 149, T20 103

The bowler took more wickets overseas (292) than at home (194), and the most in Test matches (234).

In-text Questions — Page 22Section 10.2.2 An Alternative to Pie Chart

In-text
Q1Try answering the following question based on the expenditure stacked bar chart (Fig. 10.2) you saw earlier: Which family has the smallest share of their total expenditure towards healthcare? Was it easy? If not, what visualisation can help answer such questions easily?
Answer
Family A: 1770 ÷ 8425 ≈ 21.0%
Family B: 1210 ÷ 7286 ≈ 16.6%
Family C: 1300 ÷ 7040 ≈ 18.5%

Family B has the smallest share for healthcare. It was not easy from Fig. 10.2: a “share” compares a piece with its own total, and the bars have different totals. Pie charts, or better a 100% stacked bar chart, show each family’s spending as parts of the same whole, so shares can be compared directly.

In-text Questions — Pages 23–24Section 10.2.2 Powering Up! (Example 7)

In-text
Q1The average category-wise daily electricity consumption of a house, in kWh, in 2005 and 2025 is shown in the table, with the corresponding stacked bar chart and 100% stacked bar chart.
LightingCoolingKitchen appliancesOther appliances
2005360300120220
2025220800460520
LightingCoolingKitchen appliancesOther appliances050010001500200020052025
Stacked bar chart, page 23 (daily consumption in kWh, as given in the book)
LightingCoolingKitchen appliancesOther appliances0%25%50%75%100%200536%30%12%22%202511%40%23%26%
The corresponding 100% stacked bar chart, page 24
What do you observe? What do you find interesting in the data/graphs? Discuss with others.
Answer

Some observations:

• Total use doubled, from 1000 units to 2000 units.
• Lighting fell from 360 to 220 units, and its share from 36% to 11%. More efficient bulbs (such as LEDs) could explain this.
• Cooling rose from 300 to 800 units and is now the biggest share (40%), perhaps from more air conditioners and hotter summers.
• Kitchen appliances rose almost four times (120 → 460) and their share nearly doubled (12% → 23%).

Interesting: cooling’s share rose only from 30% to 40%, but its amount rose 2.7 times. Shares hide how much the total grew; amounts hide how the mix changed. We need both charts.

Think and Reflect — Page 26Section 10.2.2 What Can We Infer? (Example 8)

Think and Reflect
Q1In Example 8, a 100% stacked bar compares the proportion of flowers blooming in Fatima’s and Naveen’s gardens across seasons. The book then shows two possible stacked bar charts (Case 1 and Case 2) that give exactly the same 100% stacked bar.
Proportion of flowers that bloomed across seasonsSummerMonsoonWinter0%25%50%75%100%Fatima50%30%20%Naveen40%25%35%
Example 8, page 24 — the 100% stacked bar chart
Number of bloomsSummerMonsoonWinterTotal
Case 1: Fatima1056342210
Case 1: Naveen483042120
Case 2: Fatima905436180
Case 2: Naveen966084240
What does this say about the scope of the stacked bars and 100% stacked bars?
Answer

A stacked bar shows actual amounts. From it we can compare totals, and parts within and across bars (for example, Fatima’s 105 summer blooms against Naveen’s 48).

A 100% stacked bar shows only proportions within each bar. It tells us that 50% of Fatima’s flowers bloomed in summer, but not how many. The same 100% chart fits Case 1 (Fatima has more summer blooms) and Case 2 (Naveen has more). So it cannot be used to compare actual numbers across bars.

Q2Given a stacked bar chart can we make a corresponding 100% stacked bar chart?
Answer

Yes. A stacked bar chart gives each part’s actual value, so we can find each bar’s total and work out every part as a percentage of it. For example, Fatima in Case 1: 105 ÷ 210 = 50%, 63 ÷ 210 = 30%, 42 ÷ 210 = 20%.

Q3Given a 100% stacked bar chart can we make a corresponding stacked bar chart?
Answer

No, not unless we are also told the total for each bar. The 100% chart gives only the percentages. Case 1 (totals 210 and 120) and Case 2 (totals 180 and 240) both give the same percentages, so we cannot tell which stacked chart it came from.

Q4What kind of inferences or comparisons can be made from a stacked bar chart and in a 100% stacked bar chart?
Answer
Stacked bar chart100% stacked bar chart
Showsactual amountspercentages (shares) of each bar’s total
Can comparetotals of different bars; parts within a bar; with care, parts across barsshares within a bar; shares across bars (which group gives a bigger fraction to a part)
Cannot tellshares easily (needs division)actual amounts or totals

For example, from Example 8’s 100% chart we can say “in Fatima’s garden more blooms came in summer than in monsoon”, but not “Fatima had more summer blooms than Naveen”.

In-text Questions — Page 26Section 10.2.2 A Typical Day (Example 9)

In-text
Q1The following chart shows the time spent on different activities in a day by the average Indian. What do you notice? What do you wonder about?
The Average Indian’s daySleepPersonal CarePaid WorkUnpaid Work & CareLearningLeisure, Social & Travel0%25%50%75%100%0h6h12h18h24h34%13%10%11%5%27%% of day
Fig. 10.5, page 26 — Time spent on different activities in a day by an average Indian (values rounded off). Source: National Time Use Survey 2024
Answer

I notice:

Sleep 34% ≈ 8.2 h; Personal care 13% ≈ 3.1 h
Paid work 10% ≈ 2.4 h; Unpaid work & care 11% ≈ 2.6 h
Learning 5% ≈ 1.2 h; Leisure, social & travel 27% ≈ 6.5 h

Sleep is the biggest part (about one-third of the day), then leisure. Paid and unpaid work take about the same time. Learning is small.

I wonder: why is learning only about 1 hour when students spend 6–7 hours in school? Why is paid work only about 2.4 hours when working adults work about 8 hours? The reason is that this is an average over all Indians — children, workers, homemakers and the elderly — so it may not describe any real person well.

Think and Reflect — Page 27Section 10.2.2 A Typical Day (Fig. 10.6)

Think and Reflect
Q1When the same data is grouped by age, we get the following stacked column chart.
Time use by age groupSleepPersonal CarePaid WorkUnpaid Work & CareLearningLeisure, Social & Travel0%25%50%75%100%0h6h12h18h24h38%13%22%26%Children (6–14)34%13%7%9%11%26%Youth (15–24)32%13%14%15%26%Adults (25–59)37%14%7%9%33%Elderly (60+)% of day
Fig. 10.6, page 27 — Age-wise grouping of the same data (values rounded off; the thin pink strip for children is about 1%). Source: National Time Use Survey 2024
What do you find interesting in the chart above? What can you infer? Discuss.
Answer

Interesting points and inferences:

• Children spend 22% (about 5.3 h) on learning and almost nothing on paid work; they also sleep the most (38%, about 9 h).
• Youth spend only 11% on learning but 7% on paid work and 9% on unpaid work: many have started working or helping at home.
• Adults spend about 29% (about 7 h) on paid plus unpaid work and almost no time learning.
• The elderly spend the most on leisure, social and travel (33%) and less on work.
• Personal care (13–14%) is almost the same for every age group.

So the “average Indian” of Fig. 10.5 hides big differences between age groups.

Q2Do you remember the sleep time over age trend that you studied last year? Does that trend align with this chart?
Answer

The usual trend is that sleep falls as people grow from childhood to adulthood: young children need the most sleep, teenagers a little less, and adults less still.

The chart mostly agrees: children 38% (about 9.1 h), youth 34% (about 8.2 h), adults 32% (about 7.7 h). The elderly (37%, about 8.9 h) do not follow the falling trend. This may be because many have retired and spend more time resting, and not because older people need more sleep.

Q3Can you explain why the learning time of people in the age group 15 – 24 has reduced significantly compared to that of the age group 6 – 14?
Answer

Learning drops from 22% (about 5.3 h) to 11% (about 2.6 h) because:

• almost all 6–14-year-olds are in school, but in the 15–24 group many have finished or left studies, especially those aged 19–24;
• many youth have started paid work (7%) or take on household and care work (9%), which children hardly do;
• the 11% is an average: full-time college students may learn as much as children, but those who are not studying bring the average down.

Q4Do adults spend about an equal amount of time in paid and unpaid work? What do you think?
Answer

On average, yes: adults spend 14% (about 3.4 h) on paid work and 15% (about 3.6 h) on unpaid work and care.

But this does not mean each adult splits the time equally. The average combines people who work long hours for pay with people (often homemakers) who do mainly unpaid household and care work. Surveys show that in India women do much more of the unpaid work and men more of the paid work. A chart split by gender would show this.

Think and Reflect — Page 28Section 10.2.2 A Typical Day

Think and Reflect
Q1Do you remember the ‘What Can A Strip Say?’ activity from last year? In each strip, if we club the tiny strips belonging to each activity together, will we get a 100% stacked bar chart like the one in Fig. 10.6?
Answer

Yes. A strip for one day has a fixed length (24 hours), with small pieces for each activity in the order they happened. If we join all the pieces of the same activity (for example, all the sleep pieces) into one block, and keep the blocks in the same order in every strip, each strip becomes one stacked bar of length 24 h = 100%.

Several such strips side by side form a 100% stacked bar chart like Fig. 10.6. We lose only the time order of the activities, not how much time each takes.

Exercise Set 10.5 — Page 29Section 10.2.2 An Alternative to Pie Chart

Q1The smart watch data of 5 people was tracked from Monday to Friday. The recorded data tells us how many hours each person spent lying down, sitting, and standing or moving around, etc., as shown in the following table.
Lying downSittingStanding or Moving around etc.
Pavani (Patient)2031
Raghu (Nurse)7413
Zakir (Teacher)8610
Sahana (Student)
Julie (___________)8106
Table 10.3, page 29 — Average number of hours spent per day (Monday – Friday)
(i) How much time does Sahana spend per day in each of these body-states? Make a reasonable guess and fill her row in the table.
(ii) Guess what activity or work Julie could be engaged in based on the data. Find out what your classmates’ guesses are.
(iii) Complete the chart below based on the tabular data.
Lying downSittingStanding/Moving around, etc.0%25%50%75%100%Pavani(Patient)Raghu(Nurse)Zakir(Teacher)Sahana(Student)Julie(________)
Chart for part (iii), page 29 — Sahana’s bar is left for you to complete
Answer

(i) Sample guess: lying down 9 h (sleep), sitting 10 h (about 6 h in class, plus homework, meals and travel), standing/moving 5 h (games, walking, chores). Total = 24 h.

(ii) Julie sits for 10 hours and moves for 6, so she probably has a desk job, such as an office worker, bank clerk or computer operator. She could also be someone who sits while working, such as a tailor or a driver.

(iii) Each value ÷ 24 × 100 gives the percentage (Sahana: 9 h = 37.5%, 10 h ≈ 41.7%, 5 h ≈ 20.8%).

Lying downSittingStanding/Moving around, etc.0%25%50%75%100%Pavani(Patient)83%13%Raghu(Nurse)29%17%54%Zakir(Teacher)33%25%42%Sahana(Student)38%42%21%Julie(________)33%42%25%
Completed chart (Sahana’s bar uses the sample guess 9 h : 10 h : 5 h)

Think and Reflect — Page 30Section 10.2.2 (after Exercise Set 10.5)

Think and Reflect
Q1Will the bars look similar if a different teacher’s data is considered instead of Zakir’s?
Answer

Roughly similar, but not the same. Most teachers sleep about 7–8 hours, stand or walk while teaching, and sit while checking work, so the bar will have a similar shape. But the details depend on the person: a teacher who teaches mostly from a chair, travels a long way by bus, or plays sport after school will have a different mix.

Q2Will the bars look different if the data for all 7 days of the week is considered?
Answer

Probably yes. Weekends are different from weekdays: people may sleep longer, students and teachers do not sit in class, a nurse may have a day off, and some people move around more doing chores or outings. The 7-day average would mix these days in, so the shares of lying down, sitting and moving would change, for example a little more lying down for most people.

Q3Some bars in the chart may also match people doing other kinds of work/activities. Can you think of any? Discuss.
Answer

Yes. For example:

• Raghu (nurse), mostly standing/moving: a shopkeeper who stands at the counter, a farmer, a delivery person, a traffic police officer, a waiter, a construction worker.
• Zakir (teacher), a mix: a salesperson, a doctor in a clinic, a chef.
• Julie, mostly sitting: an office worker, a driver, a tailor, a cashier, a software engineer.
• Pavani (patient), mostly lying down: a person recovering from an illness or injury, or a very elderly person who needs rest.

End-of-Chapter Exercises — Pages 30–36Whole chapter (questions marked * are optional/harder)

Q1In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6.
(i) In the second over they scored 12 runs. What is the run rate now?
(ii) After Over 19, their run rate was 6. What is the run rate after 20 overs given the team made 12 runs in the last over?
Answer
(i) Run rate = (6 + 12) ÷ 2 = 18 ÷ 2 = 9
(ii) Runs after 19 overs = 6 × 19 = 114
Runs after 20 overs = 114 + 12 = 126
Run rate = 126 ÷ 20 = 6.3

In (i) one big over changes the run rate a lot; in (ii) the same 12 runs raise it only a little, because it is shared over 20 overs.

Q2Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). The following is the list of the prices, in rupees (₹), that each one compiled.
Yusuf: 40(N), 105(N), 383(F), 108(N), 165(F), 116(F)
Srikanth: 194(N), 85(N), 93(N), 121(N)
Kashvi: 49(N), 297(F), 105(N), 275(F), 40(N)
Prasanna: 124(N), 333(F), 182(N), 258(F)
Gracy: 220(F), 458(F), 129(F), 183(N)
Can you describe a way they can work together to find the average price of a shuttlecock across types?
(i) Each one calculated the average of the prices they had gathered. Suppose these are ay, as, ak, ap, ag respectively. Write an expression that gives the combined average.
(ii) Suppose an, af are the average prices of the nylon shuttlecocks and the feather shuttlecocks, respectively. Write an expression that gives the average price of a shuttlecock. Will this be equal to the answer we got using the method in the previous part?
Answer

Way: each friend tells the others the total of their prices and how many prices they have (or their average and count). Add all the totals and divide by the total count, 6 + 4 + 5 + 4 + 4 = 23.

(i) Combined average = (6ay + 4as + 5ak + 4ap + 4ag) ÷ 23
(ii) There are 13 nylon and 10 feather prices:
Average = (13an + 10af) ÷ 23

Yes, the two are equal. Both numerators are the sum of all 23 prices (₹4063), so both give 4063 ÷ 23 ≈ ₹176.65.

Q3Shreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds 5 shares of the same company at an average price of ₹150.
(i) Shreyas buys 10 shares of this company at a price of ₹30 each. What is the average price per share for him after the purchase?
(ii) Vaishnavi buys some shares of this company at a price of ₹30 each and the average price per share for her after the purchase is ₹70. How many shares did she buy?
Answer
(i) Average = (25 × 150 + 10 × 30) ÷ (25 + 10)
= (3750 + 300) ÷ 35 = 4050 ÷ 35 ≈ ₹115.71

(ii) Let her buy n shares:
(5 × 150 + 30n) ÷ (5 + n) = 70
750 + 30n = 350 + 70n
400 = 40n ⇒ n = 10 shares
Q4(Pṛthūdakasvāmī, commentary on Brahmagupta’s Brāhmasphuṭasiddhānta, c. 864 CE) A “hasta” (meaning “forearm”) refers to a unit of length measuring around 18 inches.
A pool 30 hastas long is dug to different depths along its length. It is divided into 5 sections having lengths 4, 5, 6, 7, and 8 hastas, and is dug to depths of 9, 7, 7, 3, and 2 hastas, respectively. Find the mean depth of the pool.
Answer

Use Brahmagupta’s weighted mean with the section lengths as weights.

Mean depth = (4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 2) ÷ (4 + 5 + 6 + 7 + 8)
= (36 + 35 + 42 + 21 + 16) ÷ 30 = 150 ÷ 30
= 5 hastas
Q5Suvarna had purchased 1 g gold at ₹15k (short for ₹15,000). This is shown by point O denoting the average price of gold that she possesses. Six different scenarios are given below for her next transaction. For each scenario estimate and mark the average price of gold she will have after the transaction.
051015202530O
Page 31 — average price of gold (in ₹ thousand per gram); O marks ₹15k
(i) Purchase 1 g gold at ₹30k   (ii) Purchase 2 g gold at ₹30k
(iii) Purchase 1 g gold at ₹10k   (iv) Purchase 10 g gold at ₹10k
(v) Purchase 0.5 g gold at ₹30k   (vi) Purchase 0.5 g gold at ₹15k
Answer
(i) (15 + 30) ÷ 2 = 22.5k
(ii) (15 + 2 × 30) ÷ 3 = 25k
(iii) (15 + 10) ÷ 2 = 12.5k
(iv) (15 + 10 × 10) ÷ 11 ≈ 10.45k
(v) (15 + 0.5 × 30) ÷ 1.5 = 20k
(vi) (15 + 0.5 × 15) ÷ 1.5 = 15k
051015202530O(iv) 10.45(iii) 12.5(vi) 15(v) 20(i) 22.5(ii) 25
Average price after each transaction (₹ thousand per gram)
Q6Given some data with corresponding weights, how would the weighted average change if all the weights are doubled? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Answer

It does not change.

Experiment: values 10, 20 with weights 1, 3 give (10 + 60) ÷ 4 = 17.5; with weights 2, 6 they give (20 + 120) ÷ 8 = 17.5.

New average = (2w₁x₁ + 2w₂x₂ + … + 2wₙxₙ) ÷ (2w₁ + 2w₂ + … + 2wₙ)
= 2(w₁x₁ + … + wₙxₙ) ÷ 2(w₁ + … + wₙ)
= (w₁x₁ + … + wₙxₙ) ÷ (w₁ + … + wₙ) = x̄
Q7Answer the following questions based on the graph.
Cumulative number of objects currently orbiting Earth,by year of launch/separation (as of Apr. 24, 2026)Payload objects*Other objects**05,00010,00015,00020,00025,00030,00035,0001960197019801990200020102020
EoC Q7, page 32 (redrawn; bar heights are approximate, read from the printed chart). * Former transport good of a rocket (e.g., spaceship, satellite). ** Space mission components, debris, rocket stages. Source: J. McDowell, General Catalogue of Artificial Space Objects, Release 1.8.0 (2026)
(i) In 2003, approximately how many total objects were found orbiting Earth in space? In what year did this number double?
(ii) Find the approximate number and share of payload objects and other objects in the year 2025.
(iii) What can you say about the number of payload objects over time, and the number of other objects over time? What can you say about the share of payload objects over time, and the share of other objects over time?
Answer

(i) About 10,000 objects in 2003. The total reached about 20,000 (double) in about 2021; it passes 20,000 between 2021 and 2022.

(ii) In 2025 the total is about 33,000: payload ≈ 17,000 (about 51%) and other objects ≈ 16,000 (about 49%), roughly half each.

(iii) Both numbers only increase, since the chart is cumulative. Payload objects grew slowly until about 2019 (about 4,000) and then shot up to about 18,000 by 2026. Other objects grew steadily, with jumps around 2007 and 2009, and kept rising after that.
For decades payload objects were only about one-fifth to one-quarter of the total; since 2019 their share has risen sharply to about half, so the share of other objects has fallen from about three-quarters to about half.

Q8Observe the following infographic.
Jammu & Kashmir
EoC Q8, page 32 — percentage of schools in each State/UT that have a playground (redrawn on the official map of India; small States/UTs are marked with a dot; DNH & DD = Dadra & Nagar Haveli and Daman & Diu)
(i) Identify the correct inference(s) from the statements given.
(a) More schools have a playground in Haryana compared to Uttarakhand.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.
(c) Punjab has the highest number of schools with a playground.
(d) Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.
(ii) Using the information given, can we find the nation-wide percentage of schools with a playground? If not, what additional information is needed?
Answer

(i) Correct: (b) and (d).

(a) Not correct: Haryana has a higher percentage (90% vs 78%), but we do not know how many schools each state has, so we cannot compare numbers.
(b) Correct: 68% ≈ 66.7% = 2/3.
(c) Not correct: the map gives percentages, not numbers. Punjab’s 98% is the highest among the states, but Dadra & Nagar Haveli and Daman & Diu shows 99.8%.
(d) Correct: Maharashtra has both a higher percentage (93% vs 74%) and more schools, so 93% of a larger number is more than 74% of a smaller number.

(ii) No. We need the number of schools in each State/UT. The national percentage is the weighted mean of the State/UT percentages, weighted by these numbers.

Q9Decision Dilemma:
(i) Which of the plays would you choose to watch based on the following chart?
Share of ratings of three plays★★★★★★★★★★★★★★★0%25%50%75%100%Play APlay BPlay C
EoC Q9 (i), page 33 (redrawn from the printed chart; segment lengths approximate)
(ii) The corresponding stacked bar chart is shown below. Would you change your decision after looking at this chart? Why/Why not? Discuss.
Ratings of three plays★★★★★★★★★★★★★★★02505007501000Play APlay BPlay C
EoC Q9 (ii), page 33 (redrawn; number of ratings, segment lengths approximate)
Answer

(i) Play A or Play B. Each has about 71% of ratings at 4★ or 5★ and only about 14% at 1★ or 2★. Play A has a slightly bigger 5★ share, so Play A looks a touch better. Play C has the largest 5★ share (about 40%) but also about one-third 1★ ratings, so it is risky: people either love it or hate it.

(ii) Yes, I would choose Play B. The stacked chart shows that Play B has about 925 ratings, Play A only about 210 and Play C about 380. Play B’s good shares come from many more people, so they are more reliable; it is also clearly the most popular.

Q10A triathlon is an endurance race consisting of swimming 3.8 km, cycling 180 km, and running 42.2 km. Athletes compete for the fastest overall time, completing each segment in that order. The following table shows the finish time of 3 athletes in each segment.
SwimmingCyclingRunning
Athlete 101:0805:0003:15
Athlete 201:0505:1003:35
Athlete 301:2205:5503:50
Table 10.4, page 34 — Finish time of three athletes in each segment of a triathlon race (values in hh : mm)
(i) What is a suitable representation of this data — a stacked bar chart or a 100% stacked bar chart? Why do you think so?
(ii) Suppose a 100% stacked bar chart is drawn. Which of the following question(s) can be answered by looking at just that chart?
(a) Who finished the race first?
(b) Approximately what fraction of their race time did Athlete 1 spend cycling?
(c) Which athlete took the longest for running?
Answer

(i) A stacked bar chart. The race is decided by total time, and a stacked bar shows each athlete’s total as the bar length, while still showing the three segments.

Totals: Athlete 1 = 1:08 + 5:00 + 3:15 = 9 h 23 min
Athlete 2 = 1:05 + 5:10 + 3:35 = 9 h 50 min
Athlete 3 = 1:22 + 5:55 + 3:50 = 11 h 07 min

(ii) Only (b): 300 ÷ 563 ≈ 0.53, a little more than half. (a) and (c) need actual times, which a 100% chart does not show.

Q11Look at the following graph. What do you notice? What do you wonder? Write your inferences.
Distribution of disabled persons by age group andtype of disability in India (Census 2011)seeinghearingspeechmovementmental retardationmental illnessany othermultiple disability0204060801000–1920–3940–5960 & above% of disabled persons in the age group
EoC Q11, page 34 (redrawn; segment lengths are approximate, read from the printed chart). Rows are age groups.
Answer

I notice: seeing and movement disabilities form a bigger share as age increases (seeing about 18% → 25%, movement about 13% → 25%). Multiple disability is largest for 60 & above (about 12%). The shares of mental retardation (about 8% → 2%) and speech (about 9% → 4%) fall with age. Hearing stays about the same (18–20%).

Inferences: problems of eyesight, joints and walking become more common in old age, while conditions present from birth or childhood make up a larger part among the young.

I wonder: how many disabled persons are in each age group? These are only shares, so a falling share does not mean fewer people. What is included in “any other”, which is large for young people?

Q12Individual project: Do at least one of the following.
(i) Recall the previous day and fill in the approximate time spent lying down, sitting, and standing/moving around, etc. Ask at least 2 of your family members about their day and fill it in. Alternatively, you can choose to track just your own body-state for any one of the weekdays, Saturday, and Sunday. Visualise it using a stacked bar chart. Discuss with your class how you arrived at the estimated time spent in each state.
(ii) Visualise your family’s monthly expenditure using a 100% stacked bar chart.
(a) First, identify the expense categories in your family budget.
(b) Discuss with your family members to decide how you will collect and organise the data.
(c) Collect expense data for at least 3 months.
(d) Represent the data in a 100% stacked bar chart, with each bar showing how the total monthly expenditure is divided across the different categories.
(e) Finally, write your observations and inferences based on the chart.
Answer

Sample for (i): I went through the day in time blocks (sleep, school, travel, play, homework, meals) and put each block into one body-state. Each row must add to 24 h.

Weekday (sample)Lying downSittingStanding/moving
Me9 h10 h5 h
Mother7 h6 h11 h
Father7 h11 h6 h
Body-states on a weekday (sample)Lying downSittingStanding/Moving around, etc.0 h6 h12 h18 h24 hMe9105Mother7611Father7116
Sample stacked bar chart for project (i) — hours in a day

Observation: I sit the most; my mother, who does most of the housework, moves the most; my father’s office job keeps him sitting.

Q13Small-group project: Make a group of 3–4 students. Choose one scenario to design a custom rating scheme by assigning appropriate weights: (a) shopping experience at a cloth store, (b) travel experience in a bus, (c) tourism experience of a nearby tourist spot, (d) clinic/hospital experience
(i) Discuss and arrive at 4–6 aspects to rate. For each aspect chosen, justify why it matters for an overall experience.
(ii) Discuss and decide the relative weights and justify.
(iii) Collect or imagine ratings from at least 10 people on a scale of 1–5 (5 being very good).
(iv) Compute each individual rating. Compute the overall average across individuals.
(v) Make a 100% stacked bar chart using your data.
(vi) Write a short note on the observations and inferences, what your rating system captures well and what it misses.
Answer

Sample: (b) travel in a city bus.

AspectWhy it mattersWeight
Punctualitypeople plan their day around the bus3
Safetydriving and crowding affect passengers3
Cleanlinessaffects health and comfort2
Comfortseats, fans, crowding1
Staff behaviourpoliteness and help1

(iv) Person 1 rated 3, 4, 2, 3, 4: rating = (3×3 + 4×3 + 2×2 + 3×1 + 4×1) ÷ 10 = 32 ÷ 10 = 3.2. The ten individual ratings were 3.2, 3.8, 2.4, 3.8, 3.2, 4.0, 2.4, 3.4, 4.5, 3.2, so the overall average = 33.9 ÷ 10 = 3.39.

(v) For each aspect, the shares of 5★, 4★, 3★, 2★ and 1★ ratings make one 100% bar (for example, safety: 20% gave 5★, 60% gave 4★, 20% gave 3★).

(vi) The bus scores well on staff behaviour and safety but poorly on cleanliness. The scheme stresses what matters most, but it misses things like fare, frequency and crowding at peak hours.

Q14Whole class project: Each student shares the average age of their family and the number of family members. Discuss among the class and come up with a way to find out the average age of all the families of the class.
Answer

Method: for each family, multiply its average age by its number of members to get the family’s total age. Add these totals for all families and divide by the total number of family members.

Average age = (a₁n₁ + a₂n₂ + … + aₖnₖ) ÷ (n₁ + n₂ + … + nₖ)

Sample (5 families): (38.5 yr, 4), (31 yr, 3), (42 yr, 6), (29.4 yr, 5), (35 yr, 4).

Total age = 154 + 93 + 252 + 147 + 140 = 786 years
Total members = 22
Average age = 786 ÷ 22 ≈ 35.7 years
*Q15Given some data with corresponding weights, what would happen to the weighted average if all the weights are increased by a constant value, say 1? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Answer

In general it changes: it moves towards the ordinary (unweighted) average.

Experiment: values 10, 20 with weights 1, 3 give 70 ÷ 4 = 17.5. With weights 2, 4 they give 100 ÷ 6 ≈ 16.67, closer to the simple average 15.

Let W = w₁ + … + wₙ, weighted mean x̄ = (Σwᵢxᵢ) ÷ W, simple mean m = (Σxᵢ) ÷ n
New mean = Σ(wᵢ + 1)xᵢ ÷ Σ(wᵢ + 1)
= (Σwᵢxᵢ + Σxᵢ) ÷ (W + n)
= (W·x̄ + n·m) ÷ (W + n)

So the new value is a weighted mean of x̄ and m. It lies between them and is unchanged only when x̄ = m (for example, when all weights were already equal).

*Q16A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 15%. There was a decrease in the number of all three animals’ population over the year. Choose the possibilities for the change in their respective shares of the population —
(i) % of cows decreased, % of sheep decreased, % of chickens decreased
(ii) % of cows increased, % of sheep increased, % of chickens increased
(iii) % of cows remained the same, % of sheep remained the same, % of chickens remained the same
(iv) % of cows decreased, % of sheep increased, % of chickens remained the same
(v) % of cows increased, % of sheep increased, % of chickens decreased.
Answer

Possible: (iii), (iv) and (v). Not possible: (i) and (ii).

The three shares must always add to 100%, so they cannot all fall or all rise.

Examples, with last year 120 cows, 50 sheep, 30 chickens (60%, 25%, 15%):

This year (all fewer)Shares
(iii)60, 25, 1560%, 25%, 15%
(iv)55, 30, 1555%, 30%, 15%
(v)62, 28, 1062%, 28%, 10%

Chapter 10 of 14 – Maths

Also for This Chapter

NCERT Solutions Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data – Overview

ClassClass 9
SubjectMaths
Chapter10 – How Quantities Combine: Understanding Data
Textbook pagesPart II, 8 – 36
Questions solved60
Session2027
FormatFull answers on this page + PDF

NCERT Solutions Class 9 Maths Chapter 10 – FAQs

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Read every answer on this page, or download the free PDF from the table at the top.

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Yes, it follows the latest NCERT Class 9 Maths book for 2027.

How many questions are solved in NCERT Class 9 Maths Chapter 10?

All 60 questions of the chapter (textbook pages Part II, 8 – 36) – the in-text questions, activities and the end-of-chapter exercises.

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